Toán 7 Bài 4: Quy tắc dấu ngoặc và quy tắc chuyển vế Giải Toán lớp 7 trang 22 sách Chân trời sáng tạo - Tập 1

Bài trước

Mục lục

Bài sau

Giải bài tập Toán lớp 7 Bài 4: Quy tắc dấu ngoặc và quy tắc chuyển vế với lời giải chi tiết, rõ ràng theo khung chương trình sách giáo khoa Toán 7 Tập 1 Chân trời sáng tạo trang 22, 23, 24, 25. Qua đó, giúp các em ôn tập và củng cố các dạng bài tập, rèn luyện kỹ năng giải môn Toán.

Giải Toán 7 Bài 4 chi tiết phần câu hỏi, luyện tập, bài tập, đồng thời còn giúp các em hệ thống lại toàn bộ kiến thức lý thuyết trọng tâm của Bài 4 Chương I: Số hữu tỉ. Bên cạnh đó, cũng giúp thầy cô soạn giáo án cho học sinh của mình. Vậy mời thầy cô và các em cùng theo dõi bài viết dưới đây của Eballsviet.com:

Giải Toán 7 bài 4: Quy tắc dấu ngoặc và quy tắc chuyển vế Chân trời sáng tạo

Phần Hoạt động

Hoạt động 1 trang 22 Toán 7 tập 1

Tính rồi so sánh kết quả của:

a) $\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right)$ $\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right)$ và $\frac{3}{4} + \frac{1}{2} - \frac{1}{3}$ $\frac{3}{4} + \frac{1}{2} - \frac{1}{3}$

b) $\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right)$ $\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right)$ và $\frac{2}{3} - \frac{1}{2} - \frac{1}{3}$ $\frac{2}{3} - \frac{1}{2} - \frac{1}{3}$

Gợi ý đáp án:

Thực hiện các phép tính như sau:

a) Ta có:

$\begin{matrix} \dfrac{3}{4} + \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right) = \dfrac{3}{4} + \left( {\dfrac{3}{6} - \dfrac{2}{6}} \right) \hfill \\ = \dfrac{3}{4} + \dfrac{{3 - 2}}{6} = \dfrac{3}{4} + \dfrac{1}{6} = \dfrac{9}{{12}} + \dfrac{2}{{12}} = \dfrac{{11}}{{12}} \hfill \\ \end{matrix}$ $\begin{matrix} \dfrac{3}{4} + \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right) = \dfrac{3}{4} + \left( {\dfrac{3}{6} - \dfrac{2}{6}} \right) \hfill \\ = \dfrac{3}{4} + \dfrac{{3 - 2}}{6} = \dfrac{3}{4} + \dfrac{1}{6} = \dfrac{9}{{12}} + \dfrac{2}{{12}} = \dfrac{{11}}{{12}} \hfill \\ \end{matrix}$

Ta có:

$\frac{3}{4} + \frac{1}{2} - \frac{1}{3} = \frac{9}{{12}} + \frac{6}{{12}} - \frac{4}{{12}} = \frac{{9 + 6 - 4}}{{12}} = \frac{{11}}{{12}}$ $\frac{3}{4} + \frac{1}{2} - \frac{1}{3} = \frac{9}{{12}} + \frac{6}{{12}} - \frac{4}{{12}} = \frac{{9 + 6 - 4}}{{12}} = \frac{{11}}{{12}}$

Vậy $\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{3}{4} + \frac{1}{2} - \frac{1}{3}$ $\frac{3}{4} + \left( {\frac{1}{2} - \frac{1}{3}} \right) = \frac{3}{4} + \frac{1}{2} - \frac{1}{3}$

b) Ta có:

$\begin{matrix} \dfrac{2}{3} - \left( {\dfrac{1}{2} + \dfrac{1}{3}} \right) = \dfrac{2}{3} - \left( {\dfrac{3}{6} + \dfrac{2}{6}} \right) \hfill \\ = \dfrac{2}{3} - \dfrac{5}{6} = \dfrac{4}{6} - \dfrac{5}{6} = - \dfrac{1}{6} \hfill \\ \end{matrix}$ $\begin{matrix} \dfrac{2}{3} - \left( {\dfrac{1}{2} + \dfrac{1}{3}} \right) = \dfrac{2}{3} - \left( {\dfrac{3}{6} + \dfrac{2}{6}} \right) \hfill \\ = \dfrac{2}{3} - \dfrac{5}{6} = \dfrac{4}{6} - \dfrac{5}{6} = - \dfrac{1}{6} \hfill \\ \end{matrix}$

Ta có:

$\frac{2}{3} - \frac{1}{2} - \frac{1}{3} = \frac{4}{6} - \frac{3}{6} - \frac{2}{6} = \frac{{4 - 3 - 2}}{6} = \frac{{ - 1}}{6}$ $\frac{2}{3} - \frac{1}{2} - \frac{1}{3} = \frac{4}{6} - \frac{3}{6} - \frac{2}{6} = \frac{{4 - 3 - 2}}{6} = \frac{{ - 1}}{6}$

Vậy $\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right) = \frac{2}{3} - \frac{1}{2} - \frac{1}{3}$ $\frac{2}{3} - \left( {\frac{1}{2} + \frac{1}{3}} \right) = \frac{2}{3} - \frac{1}{2} - \frac{1}{3}$

Hoạt động 2 trang 23 Toán 7 tập 1

Thực hiện bài toán tìm x, biết $x - \frac{2}{5} = \frac{1}{2}$ $x - \frac{2}{5} = \frac{1}{2}$ theo hướng dẫn sau:

- Cộng hai vế với $\frac{2}{5}$ $\frac{2}{5}$

- Rút gọn hai vế;

- Ghi kết quả.

Gợi ý đáp án:

Thực hiện các phép tính như sau:

$x - \frac{2}{5} = \frac{1}{2}$ $x - \frac{2}{5} = \frac{1}{2}$

$x - \frac{2}{5} + \frac{2}{5} = \frac{1}{2} + \frac{2}{5}$ $x - \frac{2}{5} + \frac{2}{5} = \frac{1}{2} + \frac{2}{5}$(Cộng cả hai vế với 2/5)

$x + \left( { - \frac{2}{5} + \frac{2}{5}} \right) = \frac{5}{{10}} + \frac{4}{{10}}$ $x + \left( { - \frac{2}{5} + \frac{2}{5}} \right) = \frac{5}{{10}} + \frac{4}{{10}}$(Nhóm các số hạng và quy đồng mẫu số các phân số)

$x + 0 = \frac{9}{{10}}$ $x + 0 = \frac{9}{{10}}$

$x = \frac{9}{{10}}$ $x = \frac{9}{{10}}$

Vậy giá trị x cần tìm là: $x = \frac{9}{{10}}$ $x = \frac{9}{{10}}$

Phần Thực hành

Thực hành 1 trang 22 Toán 7 tập 1

Cho biểu thức:

$A = \left( {7 - \frac{2}{5} + \frac{1}{3}} \right) - \left( {6 - \frac{4}{3} + \frac{6}{5}} \right) - \left( {2 - \frac{8}{5} + \frac{5}{3}} \right)$ $A = \left( {7 - \frac{2}{5} + \frac{1}{3}} \right) - \left( {6 - \frac{4}{3} + \frac{6}{5}} \right) - \left( {2 - \frac{8}{5} + \frac{5}{3}} \right)$

Hãy tính giá trị của A bằng cách bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.

Gợi ý đáp án:

Thực hiện các phép tính như sau:

$A = 7 - \frac{2}{5} + \frac{1}{3} - 6 + \frac{4}{3} - \frac{6}{5} - 2 + \frac{8}{5} - \frac{5}{3}$ $A = 7 - \frac{2}{5} + \frac{1}{3} - 6 + \frac{4}{3} - \frac{6}{5} - 2 + \frac{8}{5} - \frac{5}{3}$----> Bỏ dấu ngoặc theo quy tắc

$A = \left( {7 - 6 - 2} \right) + \left( { - \frac{2}{5} - \frac{6}{5} + \frac{8}{5}} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right)$ $A = \left( {7 - 6 - 2} \right) + \left( { - \frac{2}{5} - \frac{6}{5} + \frac{8}{5}} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right)$---> Nhóm các hạng tử có cùng mẫu số với nhau

$\begin{matrix} A = - 1 + \dfrac{{ - 2 - 6 + 8}}{5} + \dfrac{{1 + 4 - 5}}{3} \hfill \\ A = - 1 + 0 + 0 = 1 \hfill \\ \end{matrix}$ $\begin{matrix} A = - 1 + \dfrac{{ - 2 - 6 + 8}}{5} + \dfrac{{1 + 4 - 5}}{3} \hfill \\ A = - 1 + 0 + 0 = 1 \hfill \\ \end{matrix}$

Thực hành 2 trang 23 Toán 7 tập 1

Tìm x biết:

a) $x + \frac{1}{2} = - \frac{1}{3}$ $x + \frac{1}{2} = - \frac{1}{3}$

b) $\left( { - \frac{2}{7}} \right) + x = - \frac{1}{4}$ $\left( { - \frac{2}{7}} \right) + x = - \frac{1}{4}$

Gợi ý đáp án:

Thực hiện các phép tính như sau:

a) $x + \frac{1}{2} = - \frac{1}{3}$ $x + \frac{1}{2} = - \frac{1}{3}$

$x = - \frac{1}{3} - \frac{1}{2}$ $x = - \frac{1}{3} - \frac{1}{2}$---> Chuyển số 1/2 sang vế phải và đổi dấu

$x = - \frac{2}{6} - \frac{3}{6}$ $x = - \frac{2}{6} - \frac{3}{6}$-----> Quy đồng phân số về cùng mẫu số

$\begin{matrix} x = \dfrac{{ - 6}}{6} \hfill \\ x = - 1 \hfill \\ \end{matrix}$ $\begin{matrix} x = \dfrac{{ - 6}}{6} \hfill \\ x = - 1 \hfill \\ \end{matrix}$

Vậy x = -1

b) $\left( { - \frac{2}{7}} \right) + x = - \frac{1}{4}$ $\left( { - \frac{2}{7}} \right) + x = - \frac{1}{4}$

$x = - \frac{1}{4} - \left( { - \frac{2}{7}} \right)$ $x = - \frac{1}{4} - \left( { - \frac{2}{7}} \right)$---> Chuyển số -2/7 sang vế phải và đổi dấu

$x = - \frac{1}{4} + \frac{2}{7}$ $x = - \frac{1}{4} + \frac{2}{7}$-----> Thực hiện bỏ dấu ngoặc với dấu trừ phía trước ngoặc.

$x = - \frac{7}{{28}} + \frac{8}{{28}}$ $x = - \frac{7}{{28}} + \frac{8}{{28}}$-----> Quy đồng phân số về cùng mẫu số

$x = \frac{1}{{28}}$ $x = \frac{1}{{28}}$

Vậy $x = \frac{1}{{28}}$ $x = \frac{1}{{28}}$

Thực hành 3 trang 24 Toán 7 tập 1

Tính:

a) $1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6}} \right) + \frac{1}{3}} \right]$ $1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6}} \right) + \frac{1}{3}} \right]$

b) $\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}$ $\frac{1}{3}.\left( {\frac{2}{5} - \frac{1}{2}} \right):{\left( {\frac{1}{6} - \frac{1}{5}} \right)^2}$

Gợi ý đáp án:

Thực hiện các phép tính như sau:

a) $1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6}} \right) + \frac{1}{3}} \right]$ $1\frac{1}{2} + \frac{1}{5}.\left[ {\left( { - 2\frac{5}{6}} \right) + \frac{1}{3}} \right]$

$\begin{matrix} = \dfrac{3}{2} + \dfrac{1}{5}.\left( { - \dfrac{{17}}{6} + \dfrac{1}{3}} \right) \hfill \\ = \dfrac{3}{2} + \dfrac{1}{5}.\left( { - \dfrac{{17}}{6} + \dfrac{2}{6}} \right) \hfill \\ = \dfrac{3}{2} + \dfrac{1}{5}.\left( {\dfrac{{ - 15}}{6}} \right) \hfill \\ = \dfrac{3}{2} + \left( { - \dfrac{3}{6}} \right) = \dfrac{3}{2} - \dfrac{1}{2} = \dfrac{2}{2} = 1 \hfill \\ \end{matrix}$ $\begin{matrix} = \dfrac{3}{2} + \dfrac{1}{5}.\left( { - \dfrac{{17}}{6} + \dfrac{1}{3}} \right) \hfill \\ = \dfrac{3}{2} + \dfrac{1}{5}.\left( { - \dfrac{{17}}{6} + \dfrac{2}{6}} \right) \hfill \\ = \dfrac{3}{2} + \dfrac{1}{5}.\left( {\dfrac{{ - 15}}{6}} \right) \hfill \\ = \dfrac{3}{2} + \left( { - \dfrac{3}{6}} \right) = \dfrac{3}{2} - \dfrac{1}{2} = \dfrac{2}{2} = 1 \hfill \\ \end{matrix}$

$\begin{matrix} = \dfrac{1}{3}.\left( {\dfrac{4}{{10}} - \dfrac{5}{{10}}} \right):{\left( {\dfrac{5}{{30}} - \dfrac{6}{{30}}} \right)^2} \hfill \\ = \dfrac{1}{3}.\left( { - \dfrac{1}{{10}}} \right):{\left( {\dfrac{{ - 1}}{{30}}} \right)^2} \hfill \\ = \dfrac{{ - 1}}{{30}}:{\left( {\dfrac{{ - 1}}{{30}}} \right)^2} = \left( {\dfrac{{ - 1}}{{30}}} \right):{\left( {\dfrac{{ - 1}}{{30}}} \right)^2} = \dfrac{1}{{\dfrac{{ - 1}}{{30}}}} = - 30 \hfill \\ \end{matrix}$ $\begin{matrix} = \dfrac{1}{3}.\left( {\dfrac{4}{{10}} - \dfrac{5}{{10}}} \right):{\left( {\dfrac{5}{{30}} - \dfrac{6}{{30}}} \right)^2} \hfill \\ = \dfrac{1}{3}.\left( { - \dfrac{1}{{10}}} \right):{\left( {\dfrac{{ - 1}}{{30}}} \right)^2} \hfill \\ = \dfrac{{ - 1}}{{30}}:{\left( {\dfrac{{ - 1}}{{30}}} \right)^2} = \left( {\dfrac{{ - 1}}{{30}}} \right):{\left( {\dfrac{{ - 1}}{{30}}} \right)^2} = \dfrac{1}{{\dfrac{{ - 1}}{{30}}}} = - 30 \hfill \\ \end{matrix}$

Phần Bài tập

Bài 1 trang 24 Toán 7 tập 1

Bỏ dấu ngoặc rồi tính:

a) $\left ( \frac{-3}{7}\right ) + \left ( \frac{5}{6} -\frac{4}{7} \right )$ $\left ( \frac{-3}{7}\right ) + \left ( \frac{5}{6} -\frac{4}{7} \right )$

b) $\frac{3}{5} - \left ( \frac{2}{3} +\frac{1}{5} \right )$ $\frac{3}{5} - \left ( \frac{2}{3} +\frac{1}{5} \right )$

c) $\left [ \left ( \frac{-1}{3}\right ) + 1 \right ] - \left ( \frac{2}{3} -\frac{1}{5}\right )$ $\left [ \left ( \frac{-1}{3}\right ) + 1 \right ] - \left ( \frac{2}{3} -\frac{1}{5}\right )$

d) $1\frac{1}{3} + \left (\frac{2}{3} -\frac{3}{4} \right ) - \left ( 0,8 + 1\frac{1}{5} \right )$ $1\frac{1}{3} + \left (\frac{2}{3} -\frac{3}{4} \right ) - \left ( 0,8 + 1\frac{1}{5} \right )$

Gợi ý đáp án:

a) $\left ( \frac{-3}{7}\right ) + \left ( \frac{5}{6} -\frac{4}{7} \right )$ $\left ( \frac{-3}{7}\right ) + \left ( \frac{5}{6} -\frac{4}{7} \right )$

$= \left ( \frac{-3}{7}\right ) + \left ( \frac{35}{42} -\frac{24}{42} \right )$ $= \left ( \frac{-3}{7}\right ) + \left ( \frac{35}{42} -\frac{24}{42} \right )$

$= \left ( \frac{-3}{7}\right ) + \frac{11}{42}$ $= \left ( \frac{-3}{7}\right ) + \frac{11}{42}$

$= \left ( \frac{-18}{42}\right ) + \frac{11}{42}$ $= \left ( \frac{-18}{42}\right ) + \frac{11}{42}$

$= \frac{-1}{6}$ $= \frac{-1}{6}$

b) $\frac{3}{5} - \left ( \frac{2}{3} +\frac{1}{5} \right )$ $\frac{3}{5} - \left ( \frac{2}{3} +\frac{1}{5} \right )$

$= \frac{3}{5} - \left ( \frac{10}{15} +\frac{3}{15} \right )$ $= \frac{3}{5} - \left ( \frac{10}{15} +\frac{3}{15} \right )$

$= \frac{3}{5} - \frac{13}{15}$ $= \frac{3}{5} - \frac{13}{15}$

$= \frac{9}{15} - \frac{13}{15}$ $= \frac{9}{15} - \frac{13}{15}$

$= \frac{-4}{15}$ $= \frac{-4}{15}$

c) $\left [ \left ( \frac{-1}{3}\right ) + 1 \right ] - \left ( \frac{2}{3} -\frac{1}{5}\right )$ $\left [ \left ( \frac{-1}{3}\right ) + 1 \right ] - \left ( \frac{2}{3} -\frac{1}{5}\right )$

$= \left [ \left ( \frac{-1}{3}\right ) + \frac{3}{3} \right ] - \left ( \frac{10}{15} -\frac{3}{15}\right )$ $= \left [ \left ( \frac{-1}{3}\right ) + \frac{3}{3} \right ] - \left ( \frac{10}{15} -\frac{3}{15}\right )$

$= \frac{2}{3} - \frac{7}{15}$ $= \frac{2}{3} - \frac{7}{15}$

$= \frac{10}{15} - \frac{7}{15}$ $= \frac{10}{15} - \frac{7}{15}$

$= \frac{1}{5}$ $= \frac{1}{5}$

d) $1\frac{1}{3} + \left (\frac{2}{3} -\frac{3}{4} \right ) - \left ( 0,8 + 1\frac{1}{5} \right )$ $1\frac{1}{3} + \left (\frac{2}{3} -\frac{3}{4} \right ) - \left ( 0,8 + 1\frac{1}{5} \right )$

$= \frac{4}{3} + \left (\frac{8}{12} -\frac{9}{12} \right ) - \left ( \frac{4}{5} + \frac{6}{5} \right )$ $= \frac{4}{3} + \left (\frac{8}{12} -\frac{9}{12} \right ) - \left ( \frac{4}{5} + \frac{6}{5} \right )$

$= \frac{4}{3} -\frac{1}{12} - 2$ $= \frac{4}{3} -\frac{1}{12} - 2$

$= \frac{16}{12} -\frac{1}{12} - \frac{24}{12}$ $= \frac{16}{12} -\frac{1}{12} - \frac{24}{12}$

$= -\frac{9}{12}$ $= -\frac{9}{12}$

$= -\frac{3}{4}$ $= -\frac{3}{4}$

Bài 2 trang 25 Toán 7 tập 1

Tính:

$a) \left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)$ $a) \left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)$

$b) \left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)$ $b) \left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)$

$c) \left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}$ $c) \left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}$

$d)\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]$ $d)\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]$

Gợi ý đáp án:

$\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = - 2.\end{array}$ $\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = - 2.\end{array}$

$\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ = - 2 - \frac{5}{7}.\frac{7}{{15}}\\ = - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}$ $\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ = - 2 - \frac{5}{7}.\frac{7}{{15}}\\ = - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}$

$\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}} = \frac{{ - 1}}{3}\end{array}$ $\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}} = \frac{{ - 1}}{3}\end{array}$

$\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\end{array}$ $\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\end{array}$

Bài 3 trang 25 Toán 7 tập 1

Cho biểu thức: $A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).$ $A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).$

Hãy tính giá trị của A theo hai cách:

a) Tính giá trị của từng biểu thức trong dấu ngoặc trước.

b) Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.

Gợi ý đáp án:

$\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}$ $\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}$

$\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}$ $\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}$

Bài 4 trang 25 Toán 7 tập 1

Tìm x, biết:

$a)x + \frac{3}{5} = \frac{2}{3};$ $a)x + \frac{3}{5} = \frac{2}{3};$

$b)\frac{3}{7} - x = \frac{2}{5};$ $b)\frac{3}{7} - x = \frac{2}{5};$

$c)\frac{4}{9} - \frac{2}{3}x = \frac{1}{3};$ $c)\frac{4}{9} - \frac{2}{3}x = \frac{1}{3};$

$d)\frac{3}{{10}}x - 1\frac{1}{2} = \left( {\frac{{ - 2}}{7}} \right):\frac{5}{{14}}$ $d)\frac{3}{{10}}x - 1\frac{1}{2} = \left( {\frac{{ - 2}}{7}} \right):\frac{5}{{14}}$

Gợi ý đáp án:

$\begin{array}{l}x + \frac{3}{5} = \frac{2}{3}\\x = \frac{2}{3} - \frac{3}{5}\\x = \frac{{10}}{{15}} - \frac{9}{{15}}\\x = \frac{1}{{15}}\end{array}$ $\begin{array}{l}x + \frac{3}{5} = \frac{2}{3}\\x = \frac{2}{3} - \frac{3}{5}\\x = \frac{{10}}{{15}} - \frac{9}{{15}}\\x = \frac{1}{{15}}\end{array}$

$\begin{array}{l}\frac{3}{7} - x = \frac{2}{5}\\x = \frac{3}{7} - \frac{2}{5}\\x = \frac{{15}}{{35}} - \frac{{14}}{{35}}\\x = \frac{1}{{35}}\end{array}$ $\begin{array}{l}\frac{3}{7} - x = \frac{2}{5}\\x = \frac{3}{7} - \frac{2}{5}\\x = \frac{{15}}{{35}} - \frac{{14}}{{35}}\\x = \frac{1}{{35}}\end{array}$

$\begin{array}{l}\frac{4}{9} - \frac{2}{3}x = \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} - \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} - \frac{3}{9}\\\frac{2}{3}x = \frac{1}{9}\\x = \frac{1}{9}:\frac{2}{3}\\x = \frac{1}{9}.\frac{3}{2}\\x = \frac{1}{6}\end{array}$ $\begin{array}{l}\frac{4}{9} - \frac{2}{3}x = \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} - \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} - \frac{3}{9}\\\frac{2}{3}x = \frac{1}{9}\\x = \frac{1}{9}:\frac{2}{3}\\x = \frac{1}{9}.\frac{3}{2}\\x = \frac{1}{6}\end{array}$

$\begin{array}{l}\frac{3}{{10}}x - 1\frac{1}{2} = \left( {\frac{{ - 2}}{7}} \right):\frac{5}{{14}}\\\frac{3}{{10}}x - \frac{3}{2} = \left( {\frac{{ - 2}}{7}} \right).\frac{{14}}{5}\\\frac{3}{{10}}x - \frac{3}{2} = \frac{{ - 4}}{5}\\\frac{3}{{10}}x = \frac{{ - 4}}{5} + \frac{3}{2}\\\frac{3}{{10}}x = \frac{{ - 8}}{{10}} + \frac{{15}}{{10}}\\\frac{3}{{10}}x = \frac{7}{{10}}\\x = \frac{7}{{10}}:\frac{3}{{10}}\\x = \frac{7}{3}\end{array}$ $\begin{array}{l}\frac{3}{{10}}x - 1\frac{1}{2} = \left( {\frac{{ - 2}}{7}} \right):\frac{5}{{14}}\\\frac{3}{{10}}x - \frac{3}{2} = \left( {\frac{{ - 2}}{7}} \right).\frac{{14}}{5}\\\frac{3}{{10}}x - \frac{3}{2} = \frac{{ - 4}}{5}\\\frac{3}{{10}}x = \frac{{ - 4}}{5} + \frac{3}{2}\\\frac{3}{{10}}x = \frac{{ - 8}}{{10}} + \frac{{15}}{{10}}\\\frac{3}{{10}}x = \frac{7}{{10}}\\x = \frac{7}{{10}}:\frac{3}{{10}}\\x = \frac{7}{3}\end{array}$

Bài 5 trang 25 Toán 7 tập 1

Tìm x, biết:

$a)\frac{2}{9}:x + \frac{5}{6} = 0,5;$ $a)\frac{2}{9}:x + \frac{5}{6} = 0,5;$

$b)\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3};$ $b)\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3};$

$c)1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75;$ $c)1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75;$

$d)\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}.$ $d)\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}.$

Gợi ý đáp án:

$\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}$ $\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}$

$\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}$ $\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}$

$\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}$ $\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}$

$\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}$ $\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}$

Bài 6 trang 25 Toán 7 tập 1

Tính nhanh:

$a)\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}};$ $a)\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}};$

$b) \frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}$ $b) \frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}$

$c)\left[ {\left( { - \frac{4}{9}} \right) + \frac{3}{5}} \right]:\frac{{13}}{{17}} + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}};$ $c)\left[ {\left( { - \frac{4}{9}} \right) + \frac{3}{5}} \right]:\frac{{13}}{{17}} + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}};$

$d) \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)$ $d) \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)$

Gợi ý đáp án:

$\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}$ $\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}$

$\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}$ $\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}$

$\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ = \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}$ $\begin{array}{l}\left[ {\left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ = \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}$

$\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ = - 2\end{array}$ $\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\\ = - 2\end{array}$

Chia sẻ bởi: